376 XV NORMED ALGEBRAS AND SPECTRAL THEORY Hence it is enough to show that, for each x e A, the measure mX} x is positive, or equivalently that f G(x) dmXt x(%) J> 0 for all functions G ^ 0 belonging to 3fR(Sg)» By the same reasoning, we reduce to showing that, for each y e A, we have f |j>(/)|2 dmx>x(x) *z 0; but by (15.9.2.5)* this is equivalent to g(y*yx> x) ^ 0, which can also be written g(yx, yx) S> 0 (15.6.3). This com- pletes the proof of (15.9.2.11). (15.9.2.12) Every function of the form F =£j>, where x, y e A, belongs to <D, and we have mF = mXt y . We have to show that, for all u, v in A, or equivalently (in view of (ii)) that, for all 2 e A, But the left-hand side of this relation is equal to g(zxy*u, v), and the right- hand side to g(zuv*x, y). From (15.6.3) and the commutativity of A, we have g(zxy*u, v) = g(y*zxu, v) = g(zxu, yv) and g(zuv*x, y) = g(v*zux, y) = g(zux, vy) = g(zxu, yv). Hence (15.9.2.12). We can now prove that every function G 6 tf c(Sg) belongs to $. Let K = Supp(G); then it is enough to show that there exists a function F e $ which does not vanish on K. For then we can write G = G'F, with G' 6 ^c(S^), and the lemma (15.9.2.10) will show that G e <D. For each # e K, there exists by definition an element x e A such that £(%) ^ 0, and hence a neighborhood V(x) of i in Sg such that x(%') ^ 0 for all x' e V(%). Cover K by a finite number of such neighborhoods V(#j). If xt are the corresponding elements of A, then the function F = £ xtxt does what is required, by virtue of (15.9.2.12). i (15.9.2.13) Definition ofmg and proofof(i). From (15.9.2.11), the linear form Fi-»wF(l) is a positive measure mg on Sg. Also, for each function F e <J>, we have (15.9.2.14) JWF = F-HVlinearity of the mapping FH-»/%